Word Count : 474

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We are given: * Lot size ( N = 800 ) * Sample size ( n = 20 ) * Acceptance number ( c = 3 ) * Acceptable quality ( p_a = 2% = 0.02 ) * Unacceptable quality ( p_u = 7% = 0.07 ) * Question asks for calculations at ( p = 5% = 0.05 ) We solve step by step manually. --- (i) Probability of accepting a lot when defective fraction is 5% The probability of accepting a lot in a single-sampling plan is given by the cumulative probability of observing at most ( c ) defectives in the sample: [ P_a = \sum_{x=0}^{c} \binom{n}{x} p^x (1-p)^{n-x} ] Here, ( n = 20, _________ _____ ________ _____ _________ ________.
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